[concurrency-interest] On A Formal Definition of 'Data-Race'

Nathan Reynolds nathan.reynolds at oracle.com
Wed Apr 17 12:29:48 EDT 2013


Sorry for the poor choice of method (i.e. println).  I needed something 
that would (1) take enough time to make the loop take a non-trivial 
amount of time and (2) could potentially be proved that it doesn't 
access "shared", (3) didn't have synchronization (or at least we 
wouldn't think of it as having synchronization) and (4) couldn't be 
discarded by JIT because it doesn't change global state.

Nathan Reynolds 
<http://psr.us.oracle.com/wiki/index.php/User:Nathan_Reynolds> | 
Architect | 602.333.9091
Oracle PSR Engineering <http://psr.us.oracle.com/> | Server Technology
On 4/17/2013 8:44 AM, Vitaly Davidovich wrote:
>
> I actually expect that the optimizer *would* do this transformation on 
> plain fields (provided it doesn't break intra-thread semantics, of 
> course) because it's a perf gain.
>
> Don't know how much JIT can see through println as it ultimately calls 
> into runtime and OS functions, so I'd guess it doesn't know enough or 
> simply plays conservative here.  However, Nathan's point is still 
> valid even if example isn't necessarily the best one.  If you had 
> "pure" java code instead of an I/O call that took significant time to 
> execute, the write would be delayed.  I'm not sure why that matters 
> though for benign data races.  Clearly if you need immediate 
> visibility, you code for that specifically.
>
> On Apr 17, 2013 11:32 AM, "Zhong Yu" <zhong.j.yu at gmail.com 
> <mailto:zhong.j.yu at gmail.com>> wrote:
>
>     On Wed, Apr 17, 2013 at 1:38 AM, Nathan Reynolds
>     <nathan.reynolds at oracle.com <mailto:nathan.reynolds at oracle.com>>
>     wrote:
>     > Couldn't JIT hoist the non-volatile writes out of the loop?
>
>     Certainly, sorry if my statement sounds too absolute.
>
>     > For example, the following code...
>
>     But, is this a valid example? Can JMM really reorder around
>     System.out.println()?
>
>     > for (i = 0; i < 1_000_000_000; i++)
>     > {
>     >     System.out.println(i);
>     >     shared = 2 * i;
>     > }
>     >
>     > ... could be transformed into ...
>     >
>     > for (i = 0; i < 1_000_000_000; i++)
>     > {
>     >     System.out.println(i);
>     > }
>     >
>     > shared = 2 * 1_000_000_000;
>     >
>     > ... If so, then the non-volatile write may not happen for a very
>     long time.
>     >
>     > Nathan Reynolds | Architect | 602.333.9091
>     > Oracle PSR Engineering | Server Technology
>     > On 4/16/2013 10:27 PM, Zhong Yu wrote:
>     >
>     > On Tue, Apr 16, 2013 at 8:51 PM, thurstonn
>     <thurston at nomagicsoftware.com <mailto:thurston at nomagicsoftware.com>>
>     > wrote:
>     >
>     > Vitaly Davidovich wrote
>     >
>     > The code works as-is.
>     >
>     > Absolutely.  volatile is not needed for correctness
>     >
>     > Vitaly Davidovich wrote
>     >
>     > Why?
>     >
>     > Well, for performance reasons given the 'undefined/indefinite'
>     visibility of
>     > #hash to other threads.
>     > At least according to the JMM (which has nothing to say about
>     CPU cache
>     > coherency), it is *possible* that each distinct thread that invoked
>     > #hashCode() *could* result in a recalculation of the hash.
>     >
>     > In practice though, application threads contain very frequent
>     > synchronization actions, or other operations that force VM to
>     > flush/reload. So it won't take very long for any non-volatile
>     write in
>     > one thread to become visible to other threads.
>     >
>     > Imagine a long-lived Map<String, ?>; and many threads accessing
>     the map's
>     > keyset and for some unknown reason invoking #hashCode() on each key.
>     > If #hash was declared volatile, although there is no guarantee
>     that #hash
>     > would only be calculated once, it is guaranteed that once a
>     write to main
>     > memory was completed, every *subsequent* (here meaning after the
>     write to
>     >
>     > In JMM though, we cannot even express this guarantee. Say we have
>     > threads T1...Tn, each thread Ti burns `i` seconds CPU time
>     first, then
>     > volatile-reads #hash, and if it's 0, calculates and volatile-writes
>     > #hash which takes 100 ns. We can find no guarantee from JMM that
>     > there's only one write; it's legal that every thread sees 0 from the
>     > volatile read.
>     >
>     > Zhong Yu
>     >
>     > main memory) read no matter from which thread would see #hash !=
>     0 and
>     > therefore skip the calculation.
>     >
>     >
>     >
>     > Vitaly Davidovich wrote
>     >
>     > String is too high profile (especially
>     > hashing it) to do the "naive" thing.
>     >
>     > Nothing wrong with being naive; naive can be charming.
>     >
>     >
>     > Vitaly Davidovich wrote
>     >
>     > Also, some architectures pay a
>     > penalty for volatile loads and you'd incur that each time.
>     >
>     > Fair point; the JDK authors only get one shot and they can't
>     assume that
>     > volatile reads are cheap
>     >
>     >
>     >
>     >
>     >
>     > --
>     > View this message in context:
>     >
>     http://jsr166-concurrency.10961.n7.nabble.com/On-A-Formal-Definition-of-Data-Race-tp9408p9466.html
>     > Sent from the JSR166 Concurrency mailing list archive at Nabble.com.
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